3.10 \(\int x^3 (a+b \sec (c+d x^2))^2 \, dx\)

Optimal. Leaf size=133 \[ \frac{i a b \text{PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{i a b \text{PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{d^2}+\frac{a^2 x^4}{4}-\frac{2 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac{b^2 \log \left (\cos \left (c+d x^2\right )\right )}{2 d^2}+\frac{b^2 x^2 \tan \left (c+d x^2\right )}{2 d} \]

[Out]

(a^2*x^4)/4 - ((2*I)*a*b*x^2*ArcTan[E^(I*(c + d*x^2))])/d + (b^2*Log[Cos[c + d*x^2]])/(2*d^2) + (I*a*b*PolyLog
[2, (-I)*E^(I*(c + d*x^2))])/d^2 - (I*a*b*PolyLog[2, I*E^(I*(c + d*x^2))])/d^2 + (b^2*x^2*Tan[c + d*x^2])/(2*d
)

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Rubi [A]  time = 0.164207, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {4204, 4190, 4181, 2279, 2391, 4184, 3475} \[ \frac{i a b \text{PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{i a b \text{PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )}{d^2}+\frac{a^2 x^4}{4}-\frac{2 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac{b^2 \log \left (\cos \left (c+d x^2\right )\right )}{2 d^2}+\frac{b^2 x^2 \tan \left (c+d x^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Sec[c + d*x^2])^2,x]

[Out]

(a^2*x^4)/4 - ((2*I)*a*b*x^2*ArcTan[E^(I*(c + d*x^2))])/d + (b^2*Log[Cos[c + d*x^2]])/(2*d^2) + (I*a*b*PolyLog
[2, (-I)*E^(I*(c + d*x^2))])/d^2 - (I*a*b*PolyLog[2, I*E^(I*(c + d*x^2))])/d^2 + (b^2*x^2*Tan[c + d*x^2])/(2*d
)

Rule 4204

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[
(m + 1)/n], 0] && IntegerQ[p]

Rule 4190

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x^3 \left (a+b \sec \left (c+d x^2\right )\right )^2 \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x (a+b \sec (c+d x))^2 \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (a^2 x+2 a b x \sec (c+d x)+b^2 x \sec ^2(c+d x)\right ) \, dx,x,x^2\right )\\ &=\frac{a^2 x^4}{4}+(a b) \operatorname{Subst}\left (\int x \sec (c+d x) \, dx,x,x^2\right )+\frac{1}{2} b^2 \operatorname{Subst}\left (\int x \sec ^2(c+d x) \, dx,x,x^2\right )\\ &=\frac{a^2 x^4}{4}-\frac{2 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac{b^2 x^2 \tan \left (c+d x^2\right )}{2 d}-\frac{(a b) \operatorname{Subst}\left (\int \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d}+\frac{(a b) \operatorname{Subst}\left (\int \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,x^2\right )}{d}-\frac{b^2 \operatorname{Subst}\left (\int \tan (c+d x) \, dx,x,x^2\right )}{2 d}\\ &=\frac{a^2 x^4}{4}-\frac{2 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac{b^2 \log \left (\cos \left (c+d x^2\right )\right )}{2 d^2}+\frac{b^2 x^2 \tan \left (c+d x^2\right )}{2 d}+\frac{(i a b) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{(i a b) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{i \left (c+d x^2\right )}\right )}{d^2}\\ &=\frac{a^2 x^4}{4}-\frac{2 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )}{d}+\frac{b^2 \log \left (\cos \left (c+d x^2\right )\right )}{2 d^2}+\frac{i a b \text{Li}_2\left (-i e^{i \left (c+d x^2\right )}\right )}{d^2}-\frac{i a b \text{Li}_2\left (i e^{i \left (c+d x^2\right )}\right )}{d^2}+\frac{b^2 x^2 \tan \left (c+d x^2\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.425867, size = 123, normalized size = 0.92 \[ \frac{4 i a b \text{PolyLog}\left (2,-i e^{i \left (c+d x^2\right )}\right )-4 i a b \text{PolyLog}\left (2,i e^{i \left (c+d x^2\right )}\right )+a^2 d^2 x^4-8 i a b d x^2 \tan ^{-1}\left (e^{i \left (c+d x^2\right )}\right )+2 b^2 d x^2 \tan \left (c+d x^2\right )+2 b^2 \log \left (\cos \left (c+d x^2\right )\right )}{4 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*Sec[c + d*x^2])^2,x]

[Out]

(a^2*d^2*x^4 - (8*I)*a*b*d*x^2*ArcTan[E^(I*(c + d*x^2))] + 2*b^2*Log[Cos[c + d*x^2]] + (4*I)*a*b*PolyLog[2, (-
I)*E^(I*(c + d*x^2))] - (4*I)*a*b*PolyLog[2, I*E^(I*(c + d*x^2))] + 2*b^2*d*x^2*Tan[c + d*x^2])/(4*d^2)

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Maple [F]  time = 0.291, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( a+b\sec \left ( d{x}^{2}+c \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*sec(d*x^2+c))^2,x)

[Out]

int(x^3*(a+b*sec(d*x^2+c))^2,x)

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*sec(d*x^2+c))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 2.12485, size = 1331, normalized size = 10.01 \begin{align*} \frac{a^{2} d^{2} x^{4} \cos \left (d x^{2} + c\right ) + 2 \, b^{2} d x^{2} \sin \left (d x^{2} + c\right ) - 2 i \, a b \cos \left (d x^{2} + c\right ){\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) - 2 i \, a b \cos \left (d x^{2} + c\right ){\rm Li}_2\left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) + 2 i \, a b \cos \left (d x^{2} + c\right ){\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right )\right ) + 2 i \, a b \cos \left (d x^{2} + c\right ){\rm Li}_2\left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right )\right ) -{\left (2 \, a b c - b^{2}\right )} \cos \left (d x^{2} + c\right ) \log \left (\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) +{\left (2 \, a b c + b^{2}\right )} \cos \left (d x^{2} + c\right ) \log \left (\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right ) + 2 \,{\left (a b d x^{2} + a b c\right )} \cos \left (d x^{2} + c\right ) \log \left (i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - 2 \,{\left (a b d x^{2} + a b c\right )} \cos \left (d x^{2} + c\right ) \log \left (i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) + 2 \,{\left (a b d x^{2} + a b c\right )} \cos \left (d x^{2} + c\right ) \log \left (-i \, \cos \left (d x^{2} + c\right ) + \sin \left (d x^{2} + c\right ) + 1\right ) - 2 \,{\left (a b d x^{2} + a b c\right )} \cos \left (d x^{2} + c\right ) \log \left (-i \, \cos \left (d x^{2} + c\right ) - \sin \left (d x^{2} + c\right ) + 1\right ) -{\left (2 \, a b c - b^{2}\right )} \cos \left (d x^{2} + c\right ) \log \left (-\cos \left (d x^{2} + c\right ) + i \, \sin \left (d x^{2} + c\right ) + i\right ) +{\left (2 \, a b c + b^{2}\right )} \cos \left (d x^{2} + c\right ) \log \left (-\cos \left (d x^{2} + c\right ) - i \, \sin \left (d x^{2} + c\right ) + i\right )}{4 \, d^{2} \cos \left (d x^{2} + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*sec(d*x^2+c))^2,x, algorithm="fricas")

[Out]

1/4*(a^2*d^2*x^4*cos(d*x^2 + c) + 2*b^2*d*x^2*sin(d*x^2 + c) - 2*I*a*b*cos(d*x^2 + c)*dilog(I*cos(d*x^2 + c) +
sin(d*x^2 + c)) - 2*I*a*b*cos(d*x^2 + c)*dilog(I*cos(d*x^2 + c) - sin(d*x^2 + c)) + 2*I*a*b*cos(d*x^2 + c)*di
log(-I*cos(d*x^2 + c) + sin(d*x^2 + c)) + 2*I*a*b*cos(d*x^2 + c)*dilog(-I*cos(d*x^2 + c) - sin(d*x^2 + c)) - (
2*a*b*c - b^2)*cos(d*x^2 + c)*log(cos(d*x^2 + c) + I*sin(d*x^2 + c) + I) + (2*a*b*c + b^2)*cos(d*x^2 + c)*log(
cos(d*x^2 + c) - I*sin(d*x^2 + c) + I) + 2*(a*b*d*x^2 + a*b*c)*cos(d*x^2 + c)*log(I*cos(d*x^2 + c) + sin(d*x^2
+ c) + 1) - 2*(a*b*d*x^2 + a*b*c)*cos(d*x^2 + c)*log(I*cos(d*x^2 + c) - sin(d*x^2 + c) + 1) + 2*(a*b*d*x^2 +
a*b*c)*cos(d*x^2 + c)*log(-I*cos(d*x^2 + c) + sin(d*x^2 + c) + 1) - 2*(a*b*d*x^2 + a*b*c)*cos(d*x^2 + c)*log(-
I*cos(d*x^2 + c) - sin(d*x^2 + c) + 1) - (2*a*b*c - b^2)*cos(d*x^2 + c)*log(-cos(d*x^2 + c) + I*sin(d*x^2 + c)
+ I) + (2*a*b*c + b^2)*cos(d*x^2 + c)*log(-cos(d*x^2 + c) - I*sin(d*x^2 + c) + I))/(d^2*cos(d*x^2 + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \left (a + b \sec{\left (c + d x^{2} \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*sec(d*x**2+c))**2,x)

[Out]

Integral(x**3*(a + b*sec(c + d*x**2))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x^{2} + c\right ) + a\right )}^{2} x^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*sec(d*x^2+c))^2,x, algorithm="giac")

[Out]

integrate((b*sec(d*x^2 + c) + a)^2*x^3, x)